Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
S = \frac{1}{4} + \frac{2}{{{4^2}}} + \frac{3}{{{4^3}}} + \frac{4}{{{4^4}}} + ... + \frac{{2014}}{{{4^{2014}}}}\\
\Rightarrow 4S = 1 + \frac{2}{4} + \frac{3}{{{4^2}}} + \frac{4}{{{4^3}}} + .... + \frac{{2014}}{{{4^{2013}}}}\\
\Rightarrow 4S - S = \left( {1 + \frac{2}{4} + \frac{3}{{{4^2}}} + \frac{4}{{{4^3}}} + .... + \frac{{2014}}{{{4^{2013}}}}} \right) - \left( {\frac{1}{4} + \frac{2}{{{4^2}}} + \frac{3}{{{4^3}}} + \frac{4}{{{4^4}}} + ... + \frac{{2014}}{{{4^{2014}}}}} \right)\\
\Leftrightarrow 3S = 1 + \frac{{2 - 1}}{4} + \frac{{3 - 2}}{{{4^2}}} + \frac{{4 - 3}}{{{4^3}}} + .... + \frac{{2014 - 2013}}{{{4^{2013}}}} - \frac{{2014}}{{{4^{2014}}}}\\
\Leftrightarrow 3S = 1 + \frac{1}{4} + \frac{1}{{{4^2}}} + \frac{1}{{{4^3}}} + .... + \frac{1}{{{4^{2013}}}} - \frac{{2014}}{{{4^{2014}}}}\\
A = 1 + \frac{1}{4} + \frac{1}{{{4^2}}} + \frac{1}{{{4^3}}} + .... + \frac{1}{{{4^{2013}}}}\\
\Rightarrow 4A = 4 + 1 + \frac{1}{4} + \frac{1}{{{4^2}}} + .... + \frac{1}{{{4^{2012}}}}\\
\Rightarrow 4A - A = \left( {4 + 1 + \frac{1}{4} + \frac{1}{{{4^2}}} + .... + \frac{1}{{{4^{2012}}}}} \right) - \left( {1 + \frac{1}{4} + \frac{1}{{{4^2}}} + \frac{1}{{{4^3}}} + .... + \frac{1}{{{4^{2013}}}}} \right)\\
\Leftrightarrow 3A = 4 - \frac{1}{{{4^{2013}}}}\\
\Rightarrow 3S = A - \frac{{2014}}{{{4^{2014}}}}\\
\Leftrightarrow 3S = \frac{4}{3} - \frac{1}{{{{3.4}^{2013}}}} - \frac{{2014}}{{{4^{2014}}}}\\
\Leftrightarrow S = \frac{4}{9} - \frac{1}{{{{3.4}^{2013}}}} - \frac{{2014}}{{{4^{2014}}}} < \frac{4}{9} < \frac{1}{2}
\end{array}\)
Vậy \(S < \frac{1}{2}\)
